Brothers From Different Roots
Submissions: 1411   Accuracy:

49%

  Difficulty: Easy   Marks: 2

Given two BSTs containing N1 and N2 distinct nodes respectively and given a value x. Your task is to complete the function countPairs(), that returns the count of all pairs from both the BSTs whose sum is equal to x.

Examples:

Input : BST 1:    5        
                /   \      
               3     7      
              / \   / \    
             2  4  6   8   

        BST 2:    10        
                /   \      
               6     15      
              / \   /  \    
             3  8  11  18
        x = 16
    
Output : 3
The pairs are:
(5, 11), (6, 10) and (8, 8)

Input:
The function takes three arguments as input, first the reference pointer to the root(root1) of the BST1, then reference pointer to the root(root2) of the BST2 and last the element X.
There will be T test cases and for each test case the function will be called separately.

Output:
For each test cases print the required number of pairs on new line.

Constraints:
1<=T<=100
1<=N<=103

Example:
Input:

2
7
5 3 7 2 4 6 8
7
10 6 15 3 8 11 18
16
6
10 20 30 40 5 1
5
25 35 10 15 5
30
Output:
3
2

** For More Input/Output Examples Use 'Expected Output' option **

Contributor: Harshit Sidhwa
Author: harshitsidhwa


If you have purchased any course from GeeksforGeeks then please ask your doubt on course discussion forum. You will get quick replies from GFG Moderators there.



Need help with your code? Please use ide.geeksforgeeks.org, generate link and share the link here.

to report an issue on this page.